The flux of 
is given by the surface integral
where 
is the given square region, which we can parameterize by
with 
and 
. The area element is
where 
is the normal vector to . Depending on the orientation of , this vector could be
or 
; either way, the integral reduces to
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."
Solution:
The efficiency of a Carnot cycle working between cold temperature </span>
and hot temperature
is given by
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem,
and
, the efficiency is
Therefore, the correct answer is D) 62 %.
The moon is closer to the earth than the sun
Answer:
Power=720[watt]
Explanation:
We need to remember the definition of mechanical work which is equal to the product of the force applied by the distance traveled.
In this problem, we have to find the power which is defined as the work divided into the time in which such work is performed. This way if we have the displacement and the time, this will be the speed with which this work is done.
![Power= W/T\\T=time [s]\\W=work [J]\\Power = F*V\\where\\V=velocity [m/s]\\Power=1800 * 0.4 = 720 [watt]](https://tex.z-dn.net/?f=Power%3D%20W%2FT%5C%5CT%3Dtime%20%5Bs%5D%5C%5CW%3Dwork%20%5BJ%5D%5C%5CPower%20%3D%20F%2AV%5C%5Cwhere%5C%5CV%3Dvelocity%20%5Bm%2Fs%5D%5C%5CPower%3D1800%20%2A%200.4%20%3D%20720%20%5Bwatt%5D)
Current can vary in different branches of a circuit
